POJ 2386 Lake Counting(dfs or bfs)

最新推荐文章于 2022-03-09 11:09:25 发布

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本文详细介绍了湖计数问题的解决方案，包括DFS和BFS两种方法，并提供了相应的代码实现。通过实例输入演示了如何根据给定的农场布局计算出形成的池塘数量。

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Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24999		Accepted: 12619

Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

题意：图给出的是一片农场，'W'表示水， '.' 表示土地。问这片农场有多少块池塘。

dfs解法：从任意的W开始，不停地把邻接的部分用‘.’代替。1次dfs后与初始的W连接的所有W就都被替换成了‘.’ ，

直到图中不再存在W为止，总共进行的dfs次数就是答案。

代码如下：

#include<cstdio>
int n,m;
char map[110][110];
int dir[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};

void dfs(int x,int y)
{
	int i,nowx,nowy;
	map[x][y]='.';//将W替换成‘.’ 
	for(i=0;i<8;++i)//遍历它的旁边8个方向 
	{
		nowx=x+dir[i][0];
		nowy=y+dir[i][1];
		if(nowx>=0&&nowx<n&&nowy>=0&&nowy<m&&map[nowx][nowy]=='W')
			dfs(nowx,nowy);
	}
}

int main()
{
	int i,j,cnt;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;++i)
			scanf("%s",map[i]);
		cnt=0;
		for(i=0;i<n;++i)
		{
			for(j=0;j<m;++j)
			{
				if(map[i][j]=='W')
				{
					dfs(i,j);
					cnt++;
				}
			}
		}
		printf("%d\n",cnt);
	}
	return 0;
}

bfs解法，解题思想与dfs一致，时间和内存也差不多

代码如下：

#include<cstdio>
#include<queue>
using namespace std;
#define maxn 110
char map[maxn][maxn];
int n,m;
int dir[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
struct node
{
	int x,y;
}a,temp;

void bfs(int x,int y)
{
	int i;
	queue<node>q;
	a.x=x;
	a.y=y;
	q.push(a);
	map[x][y]='.';
	while(!q.empty())
	{
		a=q.front();
		q.pop();
		for(i=0;i<8;++i)
		{
			temp.x=a.x+dir[i][0];
			temp.y=a.y+dir[i][1];
			if(temp.x>=0&&temp.x<n&&temp.y>=0&&temp.y<m&&map[temp.x][temp.y]=='W')
			{
				q.push(temp);
				map[temp.x][temp.y]='.';
			}
		} 
	}
}

int main()
{
	int i,j,cnt;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;++i)
			scanf("%s",map[i]);
		cnt=0;
		for(i=0;i<n;++i)
		{
			for(j=0;j<m;++j)
			{
				if(map[i][j]=='W')
				{
					bfs(i,j);
					cnt++;
				}
			}
		}
		printf("%d\n",cnt);
	}
	return 0;
}